Right that's where I'm having the problem
so I've created a formula by subtracting the two curves from eachother... hold on i'll put all my work on here, i thought i had gotten to an answer, but my answer didn't make sense so now i'm back at square 1

Let D(x) = | [(6 – x)^2] - [-x^2] | be the
function representing the Distance between those two given functions.
Both of the given functions are continuous.
So D(x) is continuous.
If D(x) is 0 at some point, then that’s our minimal distance.
Otherwise, D(x) is never zero.
And so it would be always strictly positive or strictly negative for any x.
Can it ever be true that those two curves cross, making D(x) = 0?
We’d like to solve –x^2 = (6-x)^2.
This gives –x^2 = 36 – 12x + x^2.
Or x^2 – 6x + 18 = 0.
This yields only complex roots, and so the curves never cross for any x.
Therefore, either D(x) is always negative, or it is always positive.
(It can’t go between being positive and negative or else it would have to be zero at some point.)
Let’s pick a test point, some easy point to evaluate D(x).
Like D(0). D(0) = 36.
So D(x) will always be positive.
Moreover, we want to minimize D(x), and we can drop the absolute value bars now.
[If D(x) was always negative, we would then want to maximize D(x) to find the minimal distance (I don’t know if that makes sense to you).]
So minimize it using calculus. Find the critical points of D’(x).
Then test to see whether those critical points are minima
(x = c is a minimum if f’(c) = 0 (or x = c is a cusp) and
f’(x) < 0 for x slightly less than c and f’(x) > 0 for x slightly bigger than c).
Then choose the smallest minimum. (I think there’s only one minimum in this case.)

I found my minimum to be x=3
but then i didn't know if that three needed to be plugged into the original equations, and then the distance found between those two points. That didn't make sense becuase the points were (3,9) and (3,-9) and the distance from that is way far